Let $\mu$ be Lebesgue measure on $[0, 1]$. Let $g : [0, 1] \to \mathbb{R}$ be Borel measurable, and set $g_{n}(x) =g(\frac{nx}{n+1})$. Assume that $g$ is bounded, and that $g$ is continuous at $x$ for $\mu$-a.e. $x$

Show by example that $\int_{0}^{1} |g_{n} − g|d\mu \to 0$ as $n \to \infty$ may fail if we drop the the hypothesis that $g$ is bounded.

my answer is as follows. could you please tell me whether it is correct or not?

consider $g(x)=\frac{1}{x}$. This function is continous a.e. hence Borel Measurable and clearly unbounded, therefore:

$\int_{0}^{1} |g_{n} − g|d\mu = \int_{0}^{1} |\frac{n+1}{nx} − \frac{1}{x}|d\mu = \frac{1}{n} \int_{0}^{1} \frac{1}{x} d\mu=+\infty$ $\forall n$

  • 1
    $\begingroup$ Yes, it is correct. $\endgroup$ – Milly Dec 3 '14 at 16:39

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